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Author: CharonChui

Algorithm/34.有效的数独.md

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+34.有效的数独
+===
+
+
+### 题目
+
+请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ，验证已经填入的数字是否有效即可。
+
+数字 1-9 在每一行只能出现一次。
+数字 1-9 在每一列只能出现一次。
+数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）
+ 
+
+注意:     
+
+- 一个有效的数独（部分已被填充）不一定是可解的。
+- 只需要根据以上规则，验证已经填入的数字是否有效即可。
+- 空白格用 '.' 表示。
+
+
+
+![image](https://raw.githubusercontent.com/CharonChui/Pictures/master/isValidSudoku.png?raw=true)
+
+
+```
+输入：board = 
+[["5","3",".",".","7",".",".",".","."]
+,["6",".",".","1","9","5",".",".","."]
+,[".","9","8",".",".",".",".","6","."]
+,["8",".",".",".","6",".",".",".","3"]
+,["4",".",".","8",".","3",".",".","1"]
+,["7",".",".",".","2",".",".",".","6"]
+,[".","6",".",".",".",".","2","8","."]
+,[".",".",".","4","1","9",".",".","5"]
+,[".",".",".",".","8",".",".","7","9"]]
+```
+输出：true
+
+```
+输入：board = 
+[["8","3",".",".","7",".",".",".","."]
+,["6",".",".","1","9","5",".",".","."]
+,[".","9","8",".",".",".",".","6","."]
+,["8",".",".",".","6",".",".",".","3"]
+,["4",".",".","8",".","3",".",".","1"]
+,["7",".",".",".","2",".",".",".","6"]
+,[".","6",".",".",".",".","2","8","."]
+,[".",".",".","4","1","9",".",".","5"]
+,[".",".",".",".","8",".",".","7","9"]]
+```
+输出：false
+解释：除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
+
+
+提示:     
+
+- board.length == 9
+- board[i].length == 9
+- board[i][j] 是一位数字（1-9）或者 '.'
+
+
+### 思路
+
+有效的数独满足以下三个条件：    
+
+- 同一个数字在每一行只能出现一次；
+
+- 同一个数字在每一列只能出现一次；
+
+- 同一个数字在每一个小九宫格只能出现一次。
+
+--- 
+
+- i是行标      
+- j是列标
+- 数字从char直接转换成int，会变成对应的ASCII数字码，而不是原来的数值。解决方法就是当前char数字减'0'：用两个char数字的ASCII码相减，差值就是原来char数值直接对应的int数值。
+    - 为什么不是减'0'，而是减'1'？
+    - 若运行了大佬的代码你会发现，减'0'的话会出现`ArrayIndexOutOfBoundsException`的异常。因为后面的代码将这个`num`作为了数组的下标。本身数组设置的就是9个位置，下标范围是`[0~8]`，那么遇到数字9作为下标的时候，不就越位了吗，所以就要减1，char转换成int的同时还能解决后面越位的情况。
+- boolean数组的巧妙建立
+    - 第一个[]存放第？行/列/块
+    - 第二个[]存放 相应数字
+    - 结合起来解释就是：第？行/列/块 是否 出现过相应数字
+
+- 行标决定一组block的起始位置（因为block为3行，所以除3取整得到组号，又因为每组block为3个，所以需要乘3），列标再细分出是哪个block（因为block是3列，所以除3取整）
+
+```
+blockIndex = i / 3 * 3 + j / 3的原因:    
+[0, 0, 0, 1, 1, 1, 2, 2, 2]
+[0, 0, 0, 1, 1, 1, 2, 2, 2]
+[0, 0, 0, 1, 1, 1, 2, 2, 2]
+[3, 3, 3, 4, 4, 4, 5, 5, 5]
+[3, 3, 3, 4, 4, 4, 5, 5, 5]
+[3, 3, 3, 4, 4, 4, 5, 5, 5]
+[6, 6, 6, 7, 7, 7, 8, 8, 8]
+[6, 6, 6, 7, 7, 7, 8, 8, 8]
+[6, 6, 6, 7, 7, 7, 8, 8, 8]
+```
+
+- blockIndex的规律探寻
+    - 微观`9x9` -> 宏观`3x3`
+    - （1）`i/3`为行号，`j/3`为列号
+    - （2）二维数组思路：`行号*列数＋列号`，即位置
+
+
+
+
+```java
+
+class Solution {
+    public boolean isValidSudoku(char[][] board) {
+        // 记录某行，某位数字是否已经被摆放
+        boolean[][] row = new boolean[9][9];
+        // 记录某列，某位数字是否已经被摆放
+        boolean[][] col = new boolean[9][9];
+        // 记录某 3x3 宫格内，某位数字是否已经被摆放
+        boolean[][] block = new boolean[9][9];
+
+        for (int i = 0; i < 9; i++) {
+            for (int j = 0; j < 9; j++) {
+                if (board[i][j] != '.') { // 只处理数字格子
+                    int num = board[i][j] - '1';
+                    int blockIndex = i / 3 * 3 + j / 3;
+                    if (row[i][num] || col[j][num] || block[blockIndex][num]) {
+                        return false;
+                    } else {
+                        row[i][num] = true;
+                        col[j][num] = true;
+                        block[blockIndex][num] = true;
+                    }
+                }
+            }
+        }
+        return true;
+    }
+}
+```
+
+---
+- 邮箱 ：charon.chui@gmail.com  
+- Good Luck! 
+
+	

Algorithm/35.螺旋矩阵.md

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+35.螺旋矩阵
+===
+
+
+### 题目
+
+
+给你一个 m 行 n 列的矩阵 matrix ，请按照 顺时针螺旋顺序 ，返回矩阵中的所有元素。
+ 
+
+示例 1:    
+![image](https://raw.githubusercontent.com/CharonChui/Pictures/master/spiralOrder_1.png?raw=true)
+
+- 输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
+- 输出：[1,2,3,6,9,8,7,4,5]
+
+示例 2:    
+![image](https://raw.githubusercontent.com/CharonChui/Pictures/master/spiralOrder_2.png?raw=true)
+
+- 输入：matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
+- 输出：[1,2,3,4,8,12,11,10,9,5,6,7]
+ 
+
+提示:    
+
+- m == matrix.length
+- n == matrix[i].length
+- 1 <= m, n <= 10
+- -100 <= matrix[i][j] <= 100
+
+
+### 思路
+
+- 对于这种螺旋遍历的方法，重要的是要确定上下左右四条边的位置
+- 初始化的时候，上边up就是0，下边down就是m-1，左边left是0，右边right是n-1。
+- 然后我们进行while循环
+    - 先遍历上边第一行up ，将所有元素加入结果res，然后上边下移一位，如果此时上边大于下边，说明此时已经遍历完成了，直接break。
+
+
+```java
+class Solution {
+    public List<Integer> spiralOrder(int[][] matrix) {
+        List<Integer> res=new ArrayList<Integer>();
+
+        if(matrix.length==0 || matrix[0].length==0) {
+            return res;
+        }
+
+        int m = matrix.length;
+        int n =matrix[0].length;
+
+        int up = 0;
+        int down = m-1;
+        int left = 0;
+        int right = n-1;
+
+        while(true) {
+            for(int i=left; i<=right; i++) {
+            	res.add(matrix[up][i]);
+            }
+
+            if (++up > down) {
+            	break;
+            }
+
+            for (int i = up; i <= down; i++) {
+            	res.add(matrix[i][right]);
+            }
+
+            if (--right < left) {
+            	break;
+            }
+
+            for (int i = right; i >= left; i--) {
+            	res.add(matrix[down][i]);
+            }
+
+            if (--down < up) {
+            	break;
+            }
+
+            for (int i = down; i >= up; i--) {
+            	res.add(matrix[i][left]);
+            }
+
+            if (++left > right) {
+            	break;
+            }
+        }
+        return res;
+    }
+}
+```
+
+---
+- 邮箱 ：charon.chui@gmail.com  
+- Good Luck! 
+
+